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Consider a deck of 52 playing cards. We draw 2 cards without replacement and let A be the event that the first card is ace and B be the event that both cards are an ace. Find B given A.

Why is the probablility of B given A not 3/51? This should be the probability of the second card you draw being an ace given the first one is an ace. I thought these probabilities would be equal but they are not the answer is 1/51.

Thanks for your help

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    $\begingroup$ I agree with the answer $\frac 3{51}$. $\endgroup$ – lulu Feb 5 '19 at 12:30
  • $\begingroup$ I think it is $3/51$, since you are drawing without replacement. Once the first ace is taken out of the deck, there are $3$ aces remaining out of a total of $51$ cards. $\endgroup$ – Micapps Feb 5 '19 at 12:32
  • $\begingroup$ ii meant 3/51, is it 3/52? $\endgroup$ – Carlos Bacca Feb 5 '19 at 12:32
  • $\begingroup$ @CarlosBacca In that case could you please edit your question so that it reads "$3/51$" rather than "$3/51$"? $\endgroup$ – Micapps Feb 5 '19 at 12:34
  • $\begingroup$ Yes, $\frac 3{51}$ is correct. $\endgroup$ – lulu Feb 5 '19 at 12:34
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It should be $\frac{3}{51}$.

$$P(B|A) = \frac{P(A\cap B)}{P(A)}= \frac{\frac{4}{52}\cdot \frac{3}{51}}{\frac{4}{52}}$$

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Here Bayes formula gives better explanation in my opinion. $$P(B/A)=\frac{P(A/B)P(B)}{P(A)}=\frac{1\cdot \frac{\binom{4}{2}\cdot \binom{48}{0}}{\binom{52}{2}}}{\frac{\binom{4}{1}\cdot \binom{48}{0}}{\binom{52}{1}}}$$ Is this right? If this is not ok I will delete it.

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